brushless-dc-motorconstant-currentcontrol
How can an electric motor, specifically a Brushless DC motor, with constant current rather than constant voltage? With the aim of having a constant torque rather than a constant voltage.
Edit: The motor will have a, more or less, constantly dynamic loading.
This is a bit off-the-cuff mathematically just to see if things stack-up: -
Your mass is 100 kg and your top speed is 20 kmph (5.56 m/s).
You want to accelerate at 0.4 m/\$s^2\$ - this means it'll take about 14 seconds to get to full speed.
The energy attained by 100 kg at 5.56 m/s is \$\dfrac{mV^2}{2}\$ = 1546 J and this equates to a power during acceleration of \$\dfrac{1546}{14}\$ = 110 watts or about 28 watts per motor.
Am I looking for something feasible by a dc brushless motor?
It definitely sounds feasible - I don't see why you shouldn't find a brushless motor like this but it'll be easier to find a brushed one for sure.
What Voltage/Amperage values am I to expect from a motor of these characteristics?
I think you can be looking at 80% efficiency so each motor will consume about 35 watts. 4 x 12V, 3A full-load rated motor should work but maybe I'd play conservative and go for something like 50 watts: -
I know these aren't cheap but it was virtually the first hit when I searched for "50W brushless motor". There should be plenty that are a bit cheaper.
Try this too - maxon are the supplier.
The easiest and best way to find the back-emf constant is to back-drive your motor with another motor and measure the voltage that is generated on an oscilloscope. Then measure the peak voltage of that wave form and divide that by the speed that you are back-driving the motor.
The units for torque constant (Kt) and back emf constant (Ke) are equivalent. The units for Kt are \$\dfrac{N \cdot m}{A}\$. If you expand that out to SI base units, you get \$\dfrac{N \cdot m}{A} = \dfrac{kg \cdot m^2}{A \cdot s^2}\$. Now when you write out the volt in SI base units you get \$V = \dfrac{kg \cdot m^2}{A \cdot s^3}\$. If you divide by \$\dfrac{rad}{s}\$, you end up with the same units as the units for Kt. So you get \$\dfrac{V}{\frac{rad}{sec}} = \dfrac{kg \cdot m^2}{A \cdot s^2} = \dfrac{N \cdot m}{A}\$.
So as you can see the units for Kt and Ke are equivalent. This equivalence holds between the torque constant and back-emf constant ONLY if we are talking about the "per phase" constants. The "per phase" constants are not usually what you find on a motor datasheet. On the datasheet, you are more concerned with how the overall torque relates to the current, not with the per phase relationships. If you work through the math, what this means is that \$Ke = Kt\$ (line-to-line) only if we are talking about an brushless motor with an ideal trapezoidal back-emf. For a motor with an ideal sinusoidal back-emf, the relationship is \$Ke = \dfrac{\sqrt{3}}{2}\cdot Kt\$ (line-to-line). In reality, brushless motors can't be made to have either ideal trapezoidal or ideal sinusoidal back-emfs. It's always somewhere in between. So usually if you measure the back-emf constant as I mentioned above it will be less than the Kt found on the datasheet.
Best Answer
The short answer- use current control. A longer answer- you have to learn how to drive a brushless motor, this is not trivial.
The common system is like that:
MCU (firmware that runs position control on top of velocity control on top of current control) creates voltage (by changing PWM duty cycle on the three phases of the motor). The feedbacks are: current from two phases, position from encoder or hall effect sensors.
The voltage command from MCU must be calculated according to current loop requirements and motor position, to take into account the motor commutation.
I know, all that sounds gibberish, but it's only because the matter itself requires many many more details to know. In the end, driving a motor with constant current is really a trivial task.
Another thing i would suggest to think about, is what will happen with constant torque. I mean, at first the motor will accelerate, at the end, if you don't have load to compensate, the torque will be zero, because the voltage will be saturated and speed will be at it's maximum.